A toy cannon uses a spring to project a 5.29-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.29-g soft rubber ball. The spring is originally compressed by 5.02 cm and has a force constant of 8.01 N/m. When the cannon is fired, the ball moves 15.7 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.0323 N on the ball.

(a) With what speed does the projectile leave the barrel of the cannon?
_________m/s

(b) At what point does the ball have maximum speed?
_________cm (from its original position)

(c) What is this maximum speed?
_________m/s

Explanation / Answer

(a)
Spring Potential Energy = 1/2*k*x^2
SP.E = 1/2*8.01 * (5.02*10^-2)^2 J
SP.E = 0.0101 J

Energy Lost in Friction, = 15.7 * 10^-2 * 0.0323 J
Energy Lost in Friction, = 0.00507 J

Using Energy Conservation,
Initial Spring Potential Energy = Energy Lost in Friction + Final K.E
0.0101 = 0.00507 + 1/2*m*v^2
0.00503 = 1/2 * 5.29 * 10^-3 * v^2
v = 1.38 m/s

(b)
Point where Spring Force > Friction Force
Fspring = Friction Force
8.01 * x = 0.0323 N
x =  0.004032 m

Distance travelled by ball, = 5.02 - 0.4032 = 4.617 cm

So ball has max speed at , x = 4.6 cm after release.

(c)
Maximum Speed,
Initial Spring Potential Energy = Energy Lost in Friction + Final K.E + Spring Potential Energy Final
1/2 * 8.01 * (5.02*10^-2)^2 = 4.6 * 10^-2 * 0.0323 + 1/2* 5.29 * 10^-3 * v^2 + 1/2 * 8.01 * (0.42*10^-2)^2
v = 1.79 m/s

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